x趋于0求（ln(1＋x²)－ln(1＋sin²x)）/xsin³x的极限

Question

x趋于0 求（ln(1＋x²)－ln(1＋sin²x)）/xsin³x的极限

Answer 1

用等价无穷小代换.
分子ln(1+x^2)-ln(1+sin(x)^2)
= ln((1+x^2)/(1+sin(x)^2))
= ln(1+(x^2-sin(x)^2)/(1+sin(x)^2)),
与(x^2-sin(x)^2)/(1+sin(x)^2)是等价无穷小(因为ln(1+t) t),
即与x^2-sin(x)^2是等价无穷小(因为1+sin(x)^2收敛到1).
而分母与x^4是等价无穷小(因为sin(x) x).
故所求极限可化为(x^2-sin(x)^2)/x^4.
之后用洛必达法则就很方便了.
或者用Taylor展开:cos(t) = 1-t^2/2+t^4/24+o(t^4),
代入得sin(x)^2 = (1-cos(2x))/2.= x^2-x^4/3+o(x^4),
故(x^2-sin(x)^2)/x^4 = 1/3+o(1)收敛到1/3.